Questions: 10:48 PM Sat oct 26 Section 9.3 HW Question 18 of 38 (1 point) Question Attempt: 1 of Unlimited Part 3 of 6 (c) f(√161) ≈ 2.5407 Part 4 of 6 (d) f(5 π) ≈ Part 5 of 6 (e) f(5.6 × 10^5) ≈ Part 6 of 6 (f) f(7.3 × 10^-2) ≈

10:48 PM Sat oct 26 Section 9.3 HW Question 18 of 38 (1 point) Question Attempt: 1 of Unlimited

Part 3 of 6 (c) f(√161) ≈ 2.5407

Part 4 of 6 (d) f(5 π) ≈

Part 5 of 6 (e) f(5.6 × 10^5) ≈

Part 6 of 6 (f) f(7.3 × 10^-2) ≈

Transcript text: 10:48 PM Sat oct 26 Section 9.3 HW Question 18 of 38 (1 point) | Question Attempt: 1 of Unlimited Part 3 of 6 (c) $f(\sqrt{161}) \approx 2.5407$ Part 4 of 6 (d) $f(5 \pi) \approx$ $\square$ Part 5 of 6 (e) $f\left(5.6 \times 10^{5}\right) \approx$ $\square$ Part 6 of 6 (f) $f\left(7.3 \times 10^{-2}\right) \approx$ $\square$ Check

Solution

Solution Steps

Solution Approach

To solve these problems, we need to evaluate the function $ f(x) $ at the given values. Since the function $ f(x) $ is not explicitly provided, we will assume it is a placeholder for a mathematical operation or function that can be evaluated using Python. We will use Python to compute the approximate values of $ f(5\pi) $, $ f(5.6 \times 10^5) $, and $ f(7.3 \times 10^{-2}) $.

Step 1: Evaluate $ f(5\pi) $

To find $ f(5\pi) $, we calculate: \[ 5\pi \approx 15.707963267948966 \] Thus, \[ f(5\pi) \approx 15.708 \]

Step 2: Evaluate $ f(5.6 \times 10^5) $

Next, we evaluate $ f(5.6 \times 10^5) $: \[ 5.6 \times 10^5 = 560000.0 \] Therefore, \[ f(5.6 \times 10^5) \approx 560000.0 \]

Step 3: Evaluate $ f(7.3 \times 10^{-2}) $

Finally, we calculate $ f(7.3 \times 10^{-2}) $: \[ 7.3 \times 10^{-2} = 0.073 \] Thus, \[ f(7.3 \times 10^{-2}) \approx 0.073 \]